A. a 3.24 m solution of base is diluted by adding water increasing its volume from 1.25 l to 3.98 l. calculate the new molarity of the solution

b. 270 ml of a 7.0 m nahco3 solution was used in the preparation of 1.9 l of diluted nahco3. what is the molarity of the diluted solution

c. what volume of 5 m naoh is needed to create a 100 ml solution of 1 m naoh?

ill get 30 extra points for these

A.   The new molarity of the diluted solution is 1.0176 M

B.   The new molarity of the diluted solution is 0.9947  M

C.   20 mL of volume is required to prepare 100ml of 1 M NaOH

Explanation:

Part A:

This problem will be solved by using the following formula

                               M₁V₁ =  M₂V₂                        (1)

Given Data:

Initial Molarity of Solution M₁ = 3.24 M

Initial volume V₁ = 1.25 L

Initial volume V₂ = 3.98 L

Final Molarity M₂= ?

Put the values in formula 1

                          3.24 M x 1.25 L = M₂ x 3.98 L

By Rearranging

                         M₂  =  3.24 M x 1.25 L / 3.98 L

                         M₂  =  4.05 / 3.98

                          M₂  =  1.0176 M

The new molarity of the diluted solution is 1.0176 M

Part B:

This problem will be solved by using the following formula

                               M₁V₁ =  M₂V₂                        (1.1)

Given Data:

Initial Molarity of Solution M₁ = 7.0 M

Initial volume V₁ = 270 mL

As the initial volume is in ml and final volume in Litre (L)

So we have to convert mL to L

       1000ml= 1L

       270 ml = ? L

       1 x 270/ 1000 = 0.27 L

So

Initial volume V₁ = 0.27 L

Initial volume V₂ = 1.9 L

Final Molarity M₂= ?

Put the values in formula 1.1

                                               M₁V₁ =  M₂V₂  

                            7.0 M x 0.27 L = M₂ x 1.9 L

By Rearranging

                         M₂  =    7.0 M x 0.27 L / 1.9 L

                         M₂  =  1.89 / 1.9

                          M₂  =  0.9947 M

The new molarity of the diluted solution is 0.9947  M

Part 3

This problem will be solved by using the following formula

                               M₁V₁ =  M₂V₂                        (1.2)

Given Data:

Initial Molarity of Solution M₁ = 5 M

Initial volume V₁ = ? mL

Initial volume V₂ =  100 mL

Final Molarity M₂= 1 M

Put the values in formula 1.2

                                               M₁V₁ =  M₂V₂  

                            5 M x V₁ = 1 M x 100 mL

By Rearranging

                         V₁  =   1 M x 100 mL/ 5 M

                         V₁  =   100 / 5

                          V₁  =   20

20 mL of volume is required to prepare 100ml of 1 M NaOH

1 a

2 d

3 a

4 c

explanation:

you need to stop taking that class